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SVEN [57.7K]
2 years ago
10

The masses of the two moons are determined to be 2M2M for Moon AA and MM for Moon BB . It is observed that the distance between

Moon BB and the planet is two times that of the distance between Moon AA and the planet. How does force exerted from the planet on Moon AA compare to the force exerted from the planet on Moon BB
Physics
2 answers:
seraphim [82]2 years ago
7 0

Answer:

 F_A = 8 F_B

Explanation:

The force exerted by the planet on each moon is given by the law of universal gravitation

        F = G \frac{m M}{r^{2} }

where M is the mass of the planet, m the mass of the moon and r the distance between its centers

let's apply this equation to our case

Moon A

the distance between the planet and the moon A is r and the mass of the moon is 2m

        F_A = G \frac{2m M}{r^{2} }

Moon B

        F_B = G \frac{m M}{(2r)^{2} }

         F_B = G \frac{m M}{4 r^{2} }

the relationship between these forces is

         F_B / F_A = \frac{1}{2 \ 4 } = 1/8

         F_A = 8 F_B

Kobotan [32]2 years ago
7 0

Answer:

F_A = 8 F_B

Explanation:

The force exerted by the planet on each moon is given by the law of universal gravitation

       F =

where M is the mass of the planet, m the mass of the moon and r the distance between its centers

let's apply this equation to our case

Moon A

the distance between the planet and the moon A is r and the mass of the moon is 2m

       F_A = G \frac{2m M}{r^{2} }

Moon B

       F_B = G \frac{m M}{(2r)^{2} }

        F_B = G \frac{m M}{4 r^{2} }

the relationship between these forces is

        F_B / F_A =  = 1/8

        F_A = 8 F_B

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A flat uniform circular disk (r= 2.00m,
dusya [7]

Incomplete question.The Complete question is here

A flat uniform circular disk (radius = 2.00 m, mass = 1.00 ✕ 102 kg) is initially stationary. The disk is free to rotate in the horizontal plane about a friction less axis perpendicular to the center of the disk. A 40.0-kg person, standing 1.25 m from the axis, begins to run on the disk in a circular path and has a tangential speed of 2.00 m/s relative to the ground.

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Explanation:

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0 = L for disk + L............... for runner

0 = Iω² - mv²r ...................they're opposite in direction

0 = (MR²/2)(ω²) - mv²r ................where is ω is angular speed which is required in part (a) of question

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0=200ω²-200

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b.)

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The person's angular speed is  

v/r = (2.00 m/s) / (1.25 m) = 1.6 rad/s

As the person and the disk are moving in opposite directions, the person will run part of a revolution and the turning disk would complete the whole revolution.

(angle) + (angle disk turns) = 2π

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t = 2.41 s

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